How do you find the antiderivative of $int (sin(sqrtx)) dx$ ?

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Answer 1 · 2017-01-26T21:55:41

$intsin(sqrtx)dx=-2sqrtxcos(sqrtx)+2sin(sqrtx)+C$

$I=intsin(sqrtx)dx$

Let $t=sqrtx$ . This implies that $x=t^2$ so $dx=2tdt$ .

Then:

$I=intsin(t)(2tdt)=int2tsin(t)dt$

Now we will use integration by parts. Let:

${(u=2t,=>,du=2dt),(dv=sin(t)dt,=>,v=-cos(t)):}$

Then:

$I=uv-intvdu$

$I=-2tcos(t)-int(-cos(t))2dt$

$I=-2tcos(t)+2intcos(t)$

$I=-2tcos(t)+2sin(t)+C$

From $t=sqrtx$ :

$I=-2sqrtxcos(sqrtx)+2sin(sqrtx)+C$

How do you find the antiderivative of int (sin(sqrtx)) dxint (sin(sqrtx)) dx?